求幂级数[∞∑n=1] [(2n-1)*x^(2n-2)]/2^n的收敛域与和函数,并求[∞∑n=1] (2n-1)/2^n的和.

S=[∞∑n=1] [(2n-1)*x^(2n-2)]/2^n
积分得：[∞∑n=1] [x^(2n-1)]/2^n
=(1/x) [∞∑n=1] [x^2/2]^n=(1/x)(x^2/2)/(1-x^2/2)=x/(2-x^2) |x^2/2|为什么要令x=1呢？谢谢。[∞∑n=1] [(2n-1)*x^(2n-2)]/2^n令x=1,刚好就是：[∞∑n=1] (2n-1)/2^n